Question 253036
Let the units, tens, and hundreds digits be U, T, and H, respectively 


Since the hundreds digits is 2 more than the tens digit, then the hundreds digit is T + 2


Therefore, the number is 100(T + 2) + 10(T) + U, or 100T + 200 + 10T + U, or 110T + U + 200


Also, the sum of the digits would be: T + 2 + T + U, or 2T + U + 2


Now, since the number (110T + U + 200) is 43 times the sum of the digits, or 2T + U + 2, then we have:


110T + U + 200 = 43(2T + U + 2)


110T + U + 200 = 86T + 43U + 86


24T – 42U = - 114


Now, since the number is divisible by 5, then its units digit, or U HAS to be either 0 or 5

Now, if we substitute 0 for U in 24T – 42U = 114, we will get T being equal to a negative number, which means that U CANNOT be 0. Therefore, U has to be 5


When 5 is substituted into the equation, 24T – 42U = -114, we get: 24T – 210 = -114, which means that 24T = 96, and T = 4


Since the tens digit, or T = 4, then the hundreds digit, or T + 2 = 6, and, as determined before, the units digit, or U = 5, which makes the 3-digit number, {{{highlight_green(645)}}}